Solve (x-2)^2-4(x+1)^2=0

$$( x - 2 ) ^ { 2 } - 4 ( x + 1 ) ^ { 2 } = 0$$

Solve for x

$$x=-4$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$x^{2}-4x+4-4\left(x+1\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}-4x+4-4\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -4 by x^{2}+2x+1.

$$x^{2}-4x+4-4x^{2}-8x-4=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-4x+4-8x-4=0$$

Combine -4x and -8x to get -12x.

$$-3x^{2}-12x+4-4=0$$

Subtract 4 from 4 to get 0.

$$-3x^{2}-12x=0$$

Factor out x.

$$x\left(-3x-12\right)=0$$

To find equation solutions, solve x=0 and -3x-12=0.

$$x=0 x=-4$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$x^{2}-4x+4-4\left(x+1\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}-4x+4-4\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -4 by x^{2}+2x+1.

$$x^{2}-4x+4-4x^{2}-8x-4=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-4x+4-8x-4=0$$

Combine -4x and -8x to get -12x.

$$-3x^{2}-12x+4-4=0$$

Subtract 4 from 4 to get 0.

$$-3x^{2}-12x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\left(-3\right)}$$

Take the square root of \left(-12\right)^{2}.

$$x=\frac{-\left(-12\right)±12}{2\left(-3\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±12}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{12±12}{-6}$$

Now solve the equation x=\frac{12±12}{-6} when ± is plus. Add 12 to 12.

$$x=\frac{24}{-6}$$

Divide 24 by -6.

$$x=-4$$

Now solve the equation x=\frac{12±12}{-6} when ± is minus. Subtract 12 from 12.

$$x=\frac{0}{-6}$$

Divide 0 by -6.

$$x=0$$

The equation is now solved.

$$x=-4 x=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$x^{2}-4x+4-4\left(x+1\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}-4x+4-4\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -4 by x^{2}+2x+1.

$$x^{2}-4x+4-4x^{2}-8x-4=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-4x+4-8x-4=0$$

Combine -4x and -8x to get -12x.

$$-3x^{2}-12x+4-4=0$$

Subtract 4 from 4 to get 0.

$$-3x^{2}-12x=0$$

Divide both sides by -3.

$$\frac{-3x^{2}-12x}{-3}=\frac{0}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{12}{-3}\right)x=\frac{0}{-3}$$

Divide -12 by -3.

$$x^{2}+4x=\frac{0}{-3}$$

Divide 0 by -3.

$$x^{2}+4x=0$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=2^{2}$$

Square 2.

$$x^{2}+4x+4=4$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+2=2 x+2=-2$$

Subtract 2 from both sides of the equation.

$$x=0 x=-4$$